Subgroups of abelian groups of order multiple of 4

Question

Why the following assertion is false?

Every abelian group of order divisible by $4$ contains a cyclic subgroup of order 4.

If the group is abelian (i.e. cyclic), then we know that cyclic groups has cyclic subgroups. I cannot see why this assertion must be false.

Lagrange's Theorem only tells us that the order of a subgroup must divide the order of a group. It doesn't say anything about the existence of such subgroups. — Sean Haight, Dec 13 '17 at 18:19
Does this answer your question? Cyclic Subgroups of Abelian Groups — tryst with freedom, Mar 10 '22 at 22:43

score 2 · Answer 1 · 2017-12-14T06:10:38.527

2

The group $\mathbb {Z}_2 \times \mathbb {Z}_2$ is not cyclic, and has order 4.

Also, Lagrange's theorem doesnt tell us anything about the existence of a subgroup of order 4.

edited Dec 14 '17 at 06:10

answered Dec 13 '17 at 18:06

Yes, but it is also not abelian, so it does not enter in the conditions of the statement. – user2820579 Dec 13 '17 at 18:12
1

It is abelian. Every group of order $p²$ with $p$ prime is abelian. – Dec 13 '17 at 18:12
See for example here: https://math.stackexchange.com/questions/443642/prove-that-every-group-of-order-4-is-abelian – Dec 13 '17 at 18:13
The smallest nonabelian group has six elements. – coffeemath Dec 13 '17 at 18:14
My mistake, Klein group is only no-cyclic. – user2820579 Dec 13 '17 at 18:15

1 Answers1